if a 100g mass is on the 60cm mark, where would a 200g mass need to be placed to balance? Also the fulcrum is at the 50cm mark

Answer 1

Let

m₁ = 100 g be one mass,

m₂ = 200 g be the balancing mass,

r₁ = 10 cm be the distance of m₁ from the fulcrum,

r₂ (to be computed) be the distance of m₂ from the fulcrum,

g be gravitational acceleration.

Balance is achieved when the net torque is zero, which is achieved

when the absolute value of the torque caused by m₁

equals the absolute value of the torque caused by m₂.

m₁gr₁ = m₂gr₂

From that

r₂ = r₁m₁/m₂

Substitute actual numbers

r₂ = 10 × 100 / 200 = 5 cm

The 200 g mass is to be placed on the 45cm mark.